Manual of Euclid. Books IV. V. VI.1859 |
Dentro del libro
Resultados 1-5 de 12
Página 6
... meet in D ; it can be proved , as in the Proposition , that DA bisects the vertical angle , and that perpendiculars DF , DG , and DE , let fall upon the sides are all equal . The circle EFG is said to be exscribed to the triangle ABC at ...
... meet in D ; it can be proved , as in the Proposition , that DA bisects the vertical angle , and that perpendiculars DF , DG , and DE , let fall upon the sides are all equal . The circle EFG is said to be exscribed to the triangle ABC at ...
Página 8
... meet in the same point , which is the centre of the circumscribing circle . Corollary 2. - If the three points of bisection of the sides of a tri- angle be joined , the perpendiculars to the sides at the points of bisec- tion are also ...
... meet in the same point , which is the centre of the circumscribing circle . Corollary 2. - If the three points of bisection of the sides of a tri- angle be joined , the perpendiculars to the sides at the points of bisec- tion are also ...
Página 71
... meet CE in E. Proof . - Because the straight line AC meets the parallels AD and EC , the angle ACE is equal to the alter- nate angle CAD ( I. 29 ) . But the angle CAD is equal to the angle BAD . There- fore the angle BAD is equal to the ...
... meet CE in E. Proof . - Because the straight line AC meets the parallels AD and EC , the angle ACE is equal to the alter- nate angle CAD ( I. 29 ) . But the angle CAD is equal to the angle BAD . There- fore the angle BAD is equal to the ...
Página 72
... meet . Let them be produced and meet in the point F. Because the angle ABC is equal 72 ELEMENTS OF EUCLID .
... meet . Let them be produced and meet in the point F. Because the angle ABC is equal 72 ELEMENTS OF EUCLID .
Página 71
... meet CE in E. Proof . Because the straight line AC meets the parallels AD and EC , the angle ACE is equal to the alter- nate angle CAD ( I. 29 ) . But the angle CAD is equal to the angle BAD . fore the angle BAD is equal to the angle ...
... meet CE in E. Proof . Because the straight line AC meets the parallels AD and EC , the angle ACE is equal to the alter- nate angle CAD ( I. 29 ) . But the angle CAD is equal to the angle BAD . fore the angle BAD is equal to the angle ...
Otras ediciones - Ver todas
Manual of Euclid: Books IV., V., VI., by J.A. Galbraith and S. Haughton Euclides Sin vista previa disponible - 2015 |
Términos y frases comunes
angle BAC angles ABC Annotation arc BC base BC is equal bisected centre circumscribed circle Corollary diameter draw drawn equal angles equal to F equi equiangular equilateral equimultiples exscribed circles find the locus fore fourth given circle given lines given point given ratio given straight lines given triangle greater ratio harmonically homologous sides inscribed intersection joining less mean proportional multiple parallel parallelogram perpendicular polygon Proof Prop Q. E. D. PROPOSITION quadrilateral radical axis radius rectangle rectilinear figure remaining angle right angle right line segments similar square Statement.-Let tangents tiple triangle ABC vertex vertical angle Wherefore αἱ ἀνάλογον δὲ δοθέντα ἐν ἔσται ἴσα ἰσάκις ἴσον καὶ κύκλον λόγον μεγέθη περὶ πλευραὶ πρὸς τὸ Πρότασις πρῶτον τὰ τὰς τὴν τῆς τοῖς τὸν αὐτὸν τοῦ τρίγωνα τῷ τῶν
Pasajes populares
Página 38 - If the first be the same multiple of the second which the third is of the fourth, and if of the first and third there be taken equimultiples, these shall be equimultiples, the one of the second, and the other of the fourth. Let A the first be the same multiple of B the second, that C the third is of D the fourth ; and of A...
Página 28 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth: or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal...
Página 3 - IN a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Página 70 - DE ; but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Página 78 - CF ; but K has to M the ratio which is compounded of the ratios of the sides ; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore equiangular parallelograms, &c.
Página 71 - Now let BD be to DC, as BA to AC, and join AD ; the angle CAD is equal to the angle DAE. The same construction being made, because BD is to DC as BA to AC ; and also BD to DC, BA to AF (2.
Página 29 - When of the equimultiples of four magnitudes (taken as in the fifth definition), the multiple of the first is greater than that of the second, but the multiple of the third is not greater than the multiple of the fourth ; then the first is said to have to the second a greater ratio than the third magnitude has to the fourth...
Página 60 - D, and other four E, F, G, H, which, two and two, have the same ratio, viz. as A is to B, so is E to F; and as B...
Página 66 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.
Página 68 - CF; and because it has been shewn, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF.